An aircraft travelling at $600\ km/h$ accelerates steadily at $10\ km/h$ per second. Taking the speed of sound as $1100\ km/h$ at the aircraft’s altitude, how long will it take to reach the ‘sound barrier’ ?

Here given, initial velocity $u=600\ km/h=600\times\frac{5}{18}=\frac{1500}{9}\ m/s$

Final velocity $v=1100\ km/h=1100\times\frac{5}{18}\ m/s=\frac{2750}{9}\ m/s$

Steady acceleration  $a=10\ km/h/s=10\frac{5}{18}\ m/s^2=\frac{25}{9}\ m/s^2$ 

On using the equation, $v=u+at$

$\frac{2750}{9}=\frac{1500}{9}+\frac{25}{9}\times t$

Or $2750=1500+25t$

Or $25t=2750-1500$

Or $25t=1250$

Or $t=\frac{1250}{25}$

Or $t=50\ seconds$

Therefore, the aircraft will take $50\ seconds$ to cross the barrier.